php 从日期中减去一定数量的小时、天、月或年
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2382458/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Subtracting a certain number of hours, days, months or years from date
提问by vitto
I'm trying to create a simple function which returns me a date with a certain number of subtracted days from now, so something like this but I dont know the date classes well:
我正在尝试创建一个简单的函数，它返回一个从现在开始减去一定天数的日期，所以是这样的，但我不太了解日期类：
<?
function get_offset_hours ($hours) {
return date ("Y-m-d H:i:s", strtotime (date ("Y-m-d H:i:s") /*and now?*/));
}
function get_offset_days ($days) {
return date ("Y-m-d H:i:s", strtotime (date ("Y-m-d H:i:s") /*and now?*/));
}
function get_offset_months ($months) {
return date ("Y-m-d H:i:s", strtotime (date ("Y-m-d H:i:s") /*and now?*/));
}
function get_offset_years ($years) {
return date ("Y-m-d H:i:s", strtotime (date ("Y-m-d H:i:s") + $years));
}
print get_offset_years (-30);
?>
Is it possible to do something similar to this? this kind of function works for years, but how to do the same with other time types?
有没有可能做类似的事情？这种功能可以工作多年，但如何对其他时间类型做同样的事情？
回答by AlexV
For hours:
用了几个小时：
function get_offset_hours($hours)
{
return date('Y-m-d H:i:s', time() + 3600 * $hours);
}
Something like that will work well for hours and days (use 86400 for days), but for months and year it's a bit trickier...
像这样的东西会在几个小时和几天内运行良好（使用 86400 几天），但对于几个月和一年来说，它有点棘手......
Also you can also do it this way:
你也可以这样做：
$date = strtotime(date('Y-m-d H:i:s') . ' +1 day');
$date = strtotime(date('Y-m-d H:i:s') . ' +1 week');
$date = strtotime(date('Y-m-d H:i:s') . ' +2 weeks');
$date = strtotime(date('Y-m-d H:i:s') . ' +1 month');
$date = strtotime(date('Y-m-d H:i:s') . ' +30 days');
$date = strtotime(date('Y-m-d H:i:s') . ' +1 year');
echo(date('Y-m-d H:i:s', $date));
回答by Phil Rykoff
Try to use datetime::sub
尝试使用 datetime::sub
Example from the docs (linked):
来自文档的示例（链接）：
<?php
$date = new DateTime("18-July-2008 16:30:30");
echo $date->format("d-m-Y H:i:s").'<br />';
date_sub($date, new DateInterval("P5D"));
echo '<br />'.$date->format("d-m-Y").' : 5 Days';
date_sub($date, new DateInterval("P5Y5M5D"));
echo '<br />'.$date->format("d-m-Y").' : 5 Days, 5 Months, 5 Years';
date_sub($date, new DateInterval("P5YT5H"));
echo '<br />'.$date->format("d-m-Y H:i:s").' : 5 Years, 5 Hours';
?>
回答by GSto
something like this:
像这样：
function offset hours($hours) {
return strtotime("+$hours hours");
}
